I bought a sheet of adapters likehttp://www.ebay.com/itm/SMD-CONVERTER-ADAPTER-PCB-SOT-TO-MSOP-SIP-DIP-14-/270832318388?pt=LH_DefaultDomain_0&hash=item3f0edd37b4Probably had like 20 each of the different adapters on it.

That looks excellent, I'll order one.

Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

From my point of view the biggest problem is not that they are pricey, it's that they are rare. Otherwise I could buy them here at home or on eBay I just talked to a friend and he says that Farnell only ships to Romania for companies, not for individuals. I'll have to check that out.

OK, I have figured it out As I wrote, one NDP6020P costs about 2.5$ at Farnell. The transport to Romania would be 5 Euro, regardless of what is in the package. So, assuming that I'd buy 15 pieces of NDP6020P (8 for the 8x8x8 cube, 5 for the 5x5x5 cube and 2 for spares), the total cost would be around 27E or 35$. That's more than the 1000 LEDs and about 33% of the total project cost I there's no other choice, I'll go for it, but if there's some more widely available, less expensive MOSFET, I'd prefer that

Yes, see http://www.aosmd.com/pdfs/datasheet/AO3401.pdf. They are rated at 4A continuous (although 3.7A would be a more realistic limit when driven from 5V, since that is the current at which Rds(on) is measured for Vgs=4.5V) and 1.4W power dissipation (although I wouldn't want to dissipate more than a few hundred mW in one).

Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

1-2 days ago we used a different calculation to obtain the value of the current limiting resistors. You wrote back then:"Resistors = inexpensive resistors, 1/8W, carbon composition. Value will depend on the LED color & the current you want to put thru them.If have a 5V source, and the anode transistor has 0.45V across it, and the cathode shif register has 0.25V across it, that leaves the remaining voltage across the LED and the resistor.For and LED with Vf of 3.2V way, and using 20mA as the current, then:(5V - .45 - .2v - 3.2)/.02 = 55 ohm. 56 ohm is a standard value."

If I understand correctly, now instead of the .45 voltage drop across the anode transistor we have the (at least) 1V drop across the IRF9540. In your earlier explanation you wrote that the cathode shift register drops 0.25V, but I think your current calculation is the correct one and it drops only about 0.07V. I use standard blue LEDs with a voltage drop of 3.3V. So, basically, if your old calculation was wrong and the new one is right, it means that the remaining voltage that needs to be handled is 5V - 1V - 0.07V - 3.3V = 0.63V. Our target current is 20 mA, so R = V/I = 1.18/0.02 = 31.5. In other words 33 Ohm resistors are fine.

Is my above calculation correct? If yes, I'll go for the IRF9540 MOSFETS. That's great news!

Hmm, are your sure you interpreted figure 2 (http://www.irf.com/product-info/datasheets/data/irf9540n.pdf) correctly? It's the first time in my life that I try to read a diagram like this, but it seems to me that the IRF9540 drops 1V at 2A. In the case of the LED cube one IRF9540 handles 64 LEDs, which means that the current going through it ranges from 0.02A (1 LED lit up) to 1.28A (64 LEDs lit up). The way I read this figure 2, at 0.02A the voltage drop is 0 but I can't figure out from the diagram how much the voltage drop is at 1.28A... I don't really understand it, but can I really use 1V as voltage drop across the IRF9540 regardless how much current (0.02A to 1.28A) is passing through it?

After looking some more at the graphs, I reached the conclusion that we should try to read figure 1, not figure 2, because figure 1 is at 25 degrees Celsius, while figure 2 is at 175 degrees Celsius. so, looking at figure 1, the second graph line from the bottom is the one for 5V. I don't really understand why the grid lines on the graph are not distributed evenly, but assuming that the distance between the 1A and 10A values on the vertical axis is distributed linearly, it seems that the voltage drop at 1.28 is aroun 0.6-0.7V and the voltage drop at 0.02A is not even represented on the drawing but it should be less than 0.3V, possibly even 0? Am I making any sense here?